Numerical computation

Scale both ratios so that the common term $b$ has the same value ($qr$) in each.

The shares add back to $N$. If only the difference between two shares is given, divide it by the difference of their ratio terms to get the value of one ratio unit.

Needs $b,d\ne 0$, and $a\ne b$ for the second form. The mean proportional of two positive numbers $a$ and $b$ is $\sqrt{ab}$; for $a\ne 0$, the third proportional to $a$ and $b$ is $\frac{b^2}{a}$ and the fourth proportional to $a$, $b$ and $c$ is $\frac{bc}{a}$ (so that $a:b=c:\frac{bc}{a}$).

Profit is shared in the ratio of capital × time when no partner draws a salary or a fixed share first. If a partner's capital changes midway, add capital × months over each period.

Positive for an increase, negative for a decrease. 'A is $x\%$ less than B' uses B as the base; the matching statement 'B is $y\%$ more than A' uses A, so $y=\frac{100x}{100-x}$ for $0<x<100$, which is larger than $x$.

Each change is taken on the value left by the previous step; write an increase as positive and a decrease as negative. For three or more changes multiply the factors: $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)\left(1+\frac{c}{100}\right)$.

Expenditure = price × consumption. If instead the price falls by $r\%$ ($r<100$), consumption can rise by $\frac{r}{100-r}\times 100\%$ for the same expenditure. The same algebra covers any product held fixed (speed × time for a fixed distance).

$p$ is the profit percentage on CP (negative for a loss); $d$ is the discount percentage on MP. Successive discounts combine like successive changes, so 20% and 10% make a single discount of $28\%$, not $30\%$.

$P$ principal, $R$ rate in per cent per annum, $T$ time in years. Interest is earned on the principal only, so the amount grows by the same sum every year (an AP).

Interest is added to the principal at the end of each period, so the amount grows by a constant factor (a GP). $CI=A-P$. For different rates in successive years multiply the factors $\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\cdots$. The same formula gives population growth at $R\%$ a year, and a value that depreciates by $R\%$ a year becomes $P{\left(1-\frac{R}{100}\right)}^T$.

Same $P$ and $R$, compounded annually.

Adding $k$ to every value adds $k$ to the average. Replacing one value changes the sum by (new − old), so the average changes by $\frac{\text{new}-\text{old}}{n}$. Work with sums, not averages, when items join or leave a group.

Quantity $q_1$ at price (or concentration) $c_1$ mixed with $q_2$ at $c_2$ gives the mean $m$. It is the weighted-average formula rearranged, so it works for prices, percentages of a component and marks. It also works for average speeds, provided the quantities are the TIMES spent at each speed, not the distances. Water counts as price 0.

A container is full with $V$ litres of pure liquid A. $x$ litres are drawn off and replaced by liquid B, and this is done $n$ times, with complete mixing each time. Each operation multiplies the amount of A by $\left(1-\frac{x}{V}\right)$, so if the container instead starts with $Q_0$ litres of A in a mixture of total volume $V$, the amount left is $Q_0{\left(1-\frac{x}{V}\right)}^n$.

Put every quantity in one unit system before calculating. For a fixed distance, time is inversely proportional to speed.

Over equal distances at speeds $u$ and $v$ the average speed is their harmonic mean, which is less than the arithmetic mean unless $u=v$.

Time to close a gap or to pass = (length to be covered) ÷ (relative speed). A train of length $L$ passes a pole or a standing person in $\frac{L}{v}$ and a platform of length $P$ in $\frac{L+P}{v}$; two trains pass each other in $\frac{L_1+L_2}{v_{\text{rel}}}$.

A starts from P towards Q and B starts from Q towards P at the same moment, each at a constant speed. After they meet, A takes time $t_A$ to reach Q and B takes time $t_B$ to reach P. Why: both travel for the same time $t$ before meeting, so $t_A=\frac{v_{B}t}{v_A}$ and $t_B=\frac{v_{A}t}{v_B}$, and dividing gives $\frac{t_B}{t_A}=\frac{v_A^2}{v_B^2}$.

$b$ is the speed in still water, $s$ the speed of the stream, $d$ the downstream and $u$ the upstream speed ($b>s$ for the boat to make headway upstream).

Rates add when workers act together; an outlet pipe, or anyone undoing the job, has a negative rate. Taking the total work as the LCM of the given times keeps every rate a whole number of units per day (or hour).

Assumes every worker has the same efficiency, so the work done is proportional to men × days × hours per day.

Valid for all real exponents when $a,b>0$; for integer exponents any non-zero $a,b$ will do. A tower is read from the top: $a^{m^n}=a^{(m^n)}$, which is not ${\left(a^m\right)}^n$.

For $a>0$, $\sqrt[q]{a}$ is the positive $q$-th root.

Solve an exponential equation by writing both sides as powers of the same base, e.g. $4^{x+1}=8^x$ gives $2^{2x+2}=2^{3x}$, so $x=2$.

For $a,b\ge 0$, with $a\ne b$ in the second. Multiply by the conjugate to clear a root from the denominator. $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$.

Find two positive numbers with sum $a$ and product $b$. For $\sqrt{a-2\sqrt{b}}$ the result is $\sqrt{x}-\sqrt{y}$ with $x>y$. Example: $\sqrt{7+2\sqrt{12}}=\sqrt{4}+\sqrt{3}=2+\sqrt{3}$.

The logarithm exists only for a positive argument and a positive base other than 1. ${\log }_{a}1=0$ and ${\log }_{a}a=1$ for every valid base. For $a>1$, ${\log }_{a}x$ increases as $x$ increases; for $0<a<1$ it decreases, so taking logarithms to such a base reverses an inequality.

For $x,y>0$. There is no law for ${\log }_a(x+y)$.

Any valid base $b$ may be used; choose 10, $e$, or whichever base makes the numbers simplest. The last form needs $k\ne 0$.

For $x,y>0$. To verify the second, take ${\log }_a$ of both sides: each becomes ${\log }_{a}x\cdot {\log }_{a}y$.

$\lfloor \cdot \rfloor$ is the greatest integer not exceeding the value. For a non-integer $N\ge 1$ the same expression counts the digits before the decimal point. For $0<N<1$, the number of zeros between the decimal point and the first non-zero digit is $\lceil -{\log }_{10}N\rceil -1$.

For an integer $a$ and a positive integer $b$, the quotient $q$ and remainder $r$ are unique. A reported remainder is never negative: a 'remainder of $-1$' on division by 7 is only a working shortcut and means 6.

Replace each number by its remainder before adding or multiplying; remainders of powers follow by repeated multiplication. $a\equiv b(\mathrm{mod}m)$ means $a$ and $b$ leave the same remainder on division by $m$.

For a prime $p$ that does not divide $a$. For the remainder of a large power on division by $p$, reduce the exponent modulo $p-1$.

$x\mid y$ reads '$x$ divides $y$'. Here $a,b$ are integers with a non-zero divisor and $n$ is a positive integer. Example: $17^{20}-1$ is divisible by both 16 and 18.

True for exactly two positive integers, not for three or more. The HCF takes the lowest power of each common prime; the LCM takes the highest power of every prime present.

Write each fraction in lowest terms first.

$p_1,\dots ,p_k$ are distinct primes. $\tau (N)$ counts all positive factors including 1 and $N$; $\sigma (N)$ is their sum. $N$ is a perfect square exactly when $\tau (N)$ is odd. $N$ can be written as a product of two positive factors (order ignored) in $\frac{\tau (N)}{2}$ ways, or $\frac{\tau (N)+1}{2}$ ways when $N$ is a perfect square. Example: $720=2^4\cdot 3^2\cdot 5$ has $5\times 3\times 2=30$ factors, so it is a product of two factors in 15 ways.

$\varphi (N)$ counts the integers from 1 to $N$ whose only common factor with $N$ is 1; $p_1,\dots ,p_k$ are the distinct primes dividing $N$. Example: $\varphi (36)=36\times \frac{1}{2}\times \frac{2}{3}=12$. Euler's theorem extends Fermat's to any modulus: if $a$ and $N$ have no common factor other than 1, then $a^{\varphi (N)}\equiv 1(\mathrm{mod}N)$; for instance $7^{12}$ leaves remainder 1 on division by 36.

Stop when $p^k>n$. The number of trailing zeros of $n!$ is this sum with $p=5$, since there are always at least as many 2s as 5s. Example: $100!$ ends in $20+4=24$ zeros.

$a$ first term, $d$ common difference, $l$ last term. The number of terms from $a$ to $l$ is $\frac{l-a}{d}+1$. Take three terms in AP as $a-d,a,a+d$ so that their sum is simply $3a$. For any sequence, not only an AP, $t_1=S_1$ and $t_n=S_n-S_{n-1}$ for $n\ge 2$.

If $r=1$, $S_n=na$. An infinite GP has a finite sum only when $|r|<1$. Take three terms in GP as $\frac{a}{r},a,ar$ so that their product is $a^3$.

Each term is an AP term times a GP term. If the formula slips from memory, write $S$, subtract $rS$ term by term and sum the GP that remains.

Equality holds only when $a=b$. A harmonic progression is a sequence whose reciprocals form an AP. AM ≥ GM also holds for $n$ positive numbers: with the sum fixed, the product is greatest when the numbers are equal, and with the product fixed, the sum is least when they are equal.

Also: the sum of the first $n$ odd numbers is $n^2$ and the sum of the first $n$ even numbers is $n(n+1)$.

Split each term into a difference so that neighbouring pieces cancel; only the first and last pieces survive.

$0\le r\le n$, no repetition. ${}^n{P}_n=n!$, ${}^n{C}_0={}^n{C}_n=1$ and $0!=1$.

If ${}^n{C}_x={}^n{C}_y$ then $x=y$ or $x+y=n$. The number of ways to select at least one item from $n$ distinct items is $2^n-1$.

For a positive integer $n$. There are $n+1$ terms and $T_{r+1}$ is the general term. Putting $x=y=1$ gives ${\sum }_{r=0}^n{}^n{C}_r=2^n$.

Anything counted twice is subtracted once, and so on. Among $1,2,\dots ,N$ exactly $\lfloor \frac{N}{k}\rfloor$ numbers are multiples of $k$. Example: from 1 to 100, the numbers divisible by 2 or 3 number $50+33-16=67$ (16 being the multiples of 6).

$r$ positions each filled from $n$ distinct kinds in unlimited supply give $n^r$ ordered arrangements. When only how many of each kind are chosen matters, the number of selections of $r$ items from $n$ kinds is ${}^{n+r-1}{C}_r$.

$n$ objects of which $p_1$ are alike of one kind, $p_2$ alike of another, and so on ($p_1+p_2+\cdots +p_k=n$; an object with no duplicate counts as a kind with $p=1$). Selecting at least one item from $p$ identical items of one kind, $q$ of a second kind and $t$ of a third can be done in $(p+1)(q+1)(t+1)-1$ ways.

For $n$ distinct objects only the relative order counts. The halved count ($n\ge 3$) applies to necklaces, garlands and rings of beads that can be turned over, never to people seated round a table. If the seats are numbered, or one position is otherwise marked, the arrangement is linear: $n!$.

Equivalently, $x_1+x_2+\cdots +x_r=n$ has ${}^{n+r-1}{C}_{r-1}$ solutions in non-negative integers and ${}^{n-1}{C}_{r-1}$ in positive integers (needs $n\ge r$). The count $r^n$ for distinct objects allows empty boxes; with no box empty it is $r^n-{}^r{C}_1(r-1{)}^n+{}^r{C}_2(r-2{)}^n-\cdots$, by inclusion and exclusion. Example: 4 distinct objects into 3 boxes with none empty: $81-3\times 16+3\times 1=36$.

Ways to split $mn$ distinct objects into $m$ unlabelled groups of $n$ each. Drop the $m!$ when the groups are labelled (named teams, rooms, or people receiving them). Example: 4 players into two pairs: $\frac{4!}{(2!{)}^{2}2!}=3$. For groups of different sizes $p$, $q$ and $s$ no division by the number of groups is needed: $\frac{(p+q+s)!}{p!q!s!}$, so 9 objects split into groups of 2, 3 and 4 in $\frac{9!}{2!3!4!}=1260$ ways.

Arrangements of $n$ distinct items in which none is in its own original place (letters in wrong envelopes). $D_1=0$, $D_2=1$, $D_3=2$, $D_4=9$, $D_5=44$.

Lines and triangles are for $n$ points with no three collinear. If $m$ of them ($m\ge 3$) lie on one line and no other three are collinear, subtract ${}^m{C}_2-1$ lines and ${}^m{C}_3$ triangles. Grid paths counts the shortest routes made of $m$ unit steps right and $n$ unit steps up in any order.

Good when $|nx|$ is a few per cent; the neglected term is about $\frac{n(n-1)}{2}{x}^2$. Example: $1.02^5\approx 1.10$ (exact value $1.1041$).

Example: $\sqrt{1604}\approx 40+\frac{4}{80}=40.05$; indeed $40.05^2=1604.0025$.

For small relative changes. A 2% rise in length and a 3% rise in width raise the area by about 5% (exactly $5.06\%$).

An approximation for annual compounding at a moderate rate of $R\%$ per year; the exact time is $\frac{\ln 2}{\ln \left(1+\frac{R}{100}\right)}$. At 8%: estimate 9 years, exact about 9.01 years.